Question: Find the general solution of

$$x^2 (y-xy') =y(y') ^2$$

if the singular solution doesn't exist.

Now, I know that it has to be solved by Clairaut's equation.

However, the given equation is not of the form $y=px+f(p)$ where $p=y'$ and cannot, as far as I know, be reduced to this form.

I even tried arranging the equation in the form of a quadratic in $y'$ but when I find the solution it yields a rather big equation that doesn't really lead anywhere.

Please let me know if you have any suggestions.



Rewrite the equation as $$x^2 \left(y-\frac{x}{x'}\right)-\frac{y}{x'^2}=0$$ Now, let $x=\sqrt{y z}$ which makes $$\frac{y^2 z \left(y^2 z'^2-z^2-4\right)}{\left(y z'+z\right)^2}=0$$ Excluding the trivial solution, we are left with $$y^2 z'^2-z^2-4=0$$ that is to say $$\frac{z'^2}{z^2+4}=\frac 1 {y^2}\implies \frac{z'}{\sqrt{z^2+4}}=\pm \frac 1y$$ which seems to be simple to integrate.

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  • $\begingroup$ Shouldn't the denominator be yz'+zy' when you substitute for x or am I missing something here? $\endgroup$ – MadCap Nov 28 '18 at 0:27
  • $\begingroup$ @MadCap. Since $y$ became the variable, $y'=1$. I double check and all of this seems to be correct. $\endgroup$ – Claude Leibovici Nov 28 '18 at 3:08
  • $\begingroup$ Oh now it's more clear. $\endgroup$ – MadCap Nov 28 '18 at 13:18

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